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When the radius of the circumscribed circle of △ ABC is 1, the opposite sides of angles a, B and C are a, B and C respectively, and the angles a, B and C form an arithmetic sequence. The value range of A2 + C2 is calculated
From a, B, C into the arithmetic sequence, we know that B = 60 degrees
From the sine theorem there is a
sinA=b
sinB=c
sinC=2R，
There is b = 2rsinb = 2 × 1 ×
Three
2=
3，
That is, B2 = A2 + c2-2acccosb = A2 + c2-2ac × 1
2=a2+c2-ac．
That is, A2 + C2 = B2 + AC > 3
And there is A2 + C2 = B2 + AC ≤ 3 + A2 + C2
c，
So A2 + C2 ≤ 6, that is, the range of A2 + C2 is (3,6]
When the radius of the circumscribed circle of ABC is 1, the angle A.B.C is an arithmetic sequence, and the length of the side to which the angle A.B.C is opposite is a.b.c. find out the value range of a ^ 2 + C ^ 2
From 2B = a + C
A + B + C = 180 degrees
B = 60 degrees
Let SINB = 2 be the circumcircle
So B = 2sinb = 2sin60 = root 3
By cosine theorem
a^2+c^2=b^2+2ac*cos60=3+ac
When the radius of the circumscribed circle of △ ABC is 1, the opposite sides of angles a, B and C are a, B and C respectively, and the angles a, B and C form an arithmetic sequence. The value range of A2 + C2 is calculated
According to the sine theorem, there is a = bsinb = csinc = 2R, B = 2rsinb = 2 × 1 × 32 = 3, that is, B2 = A2 + c2-2acccosb = A2 + c2-2ac × 12 = A2 + c2-ac
In the triangle ABC, if the three angles a, B, C form an arithmetic sequence, and the edge B = 2, then the area of the circumscribed circle is
Make two auxiliary lines, Ao and Bo B = 60, AOB = 120, B = 2 in the isosceles triangle AOB, knowing B and B, we can find the waist, that is, r = two-thirds of the root three, and the area is 4 π / 3
In the triangle ABC, a, B, C are the opposite sides of angle ABC respectively. If a = 2B COSC, then the triangle must be
By cosine theorem
cosC=(a^2+b^2-c^2)/(2ab)
It is known that a = 2B * COSC = 2B * (a ^ 2 + B ^ 2-C ^ 2) / (2Ab) = (a ^ 2 + B ^ 2-C ^ 2) / A
B ^ 2-C ^ 2 = 0
B and C are all triangular sides, so b > 0, C > 0
Continue to simplify and get b = C
So this triangle must be an isosceles triangle
In the triangle ABC, ab = 2, AC = radical 3, angle a = angle BCD = 45 degrees, D is on the extension line of AB, calculate the area of BC and triangle ABC And the area of BCD
Because a / Sina = B / SINB = C / SINB, then it can be obtained by two formulas. In triangle ADC and triangle CBD. And triangle ABC. Respectively, there is a key condition ad = AB + BD. from this, we can get BC by combining three trigonometric forms, then we can get the area product for a long time
In the triangle ABC, ∠ ACB = 90 °, ab = radical 8, BC = radical 2, to find the height CD on the hypotenuse
According to Pythagorean theorem, AC = radical 6
S △ ABC = AB * CD / 2 = AC * BC / 2 ℅ CD = AC * BC / AB = root 12 / root 8 = root 6 / 2
If PA is perpendicular to plane ABC, AC is perpendicular to BC, PA = AC = 1, BC = root 2, find cosine value of dihedral angle a-pb-c Come back quickly
∵ PA is perpendicular to the plane ABC, ∵ PA ⊥ AC, PA ⊥ ab
In RT Δ PAC, PC = √ (1? + 1?) = √ 2
In RT Δ ACB, ab = √ [1] = √ 3
In RT Δ BAP, BP = √ [1] + (√ 3) 2] = 2, and ∵ PC = BC = √ 2
The Δ PCB is an isosceles right triangle
Take the midpoint e of Pb, connect CE, CE = PE = 1;
Connection AE, AE = Pb / 2 = 1
The Δ AEC is an equilateral triangle
The cosine value of dihedral angle a-pb-c is 1 / 2
In p-abc of a triangular pyramid, if the vertical plane ABC AC of PA is vertical BC AB = 2 BC = radical 2 Pb = radical 6, then the size of dihedral angle p-bc-a is
∵ PA ⊥ plane ABC, BC ⊥ AC,
According to the three perpendicular line theorem,
BC⊥PC,
Qi
As shown in the figure, in the pyramid p-abc, PA = Pb = PC = AC = 4, ab = BC = 2, radical 2 (1) Verification: plane ABC vertical plane APC (2) to find the sine value of the angle between the straight line PA and the plane PBC
The first problem: take the midpoint of AC as D. ∵ AB = BC = 2 √ 2, AC = 4, ᙽ AB ^ 2 + BC ^ 2 = AC ^ 2,
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